The charge on a parallel plate capacitor varies as $q = q_0\cos 2\pi\nu t$. The plates are very large and close together (area $= A$, separation $= d$). Neglecting the edge effects, find the displacement current through the capacitor.

The charge on a parallel plate capacitor varies as q = q₀cos 2πnu t. The plates are very large and close toget

NCERT Exemplar · Q10

q = q_0\cos 2\pi\nu t

Karnataka PUCSubjective· 2mImportance★★★★★

48% · 22/46 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Displacement Current

Displacement Current

The Problem Maxwell Spotted

Ampere's circuital law, in its original form, links the magnetic field around a closed loop to the conduction current (moving charges) threading that loop:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c

Maxwell realised this law is incomplete. The classic illustration is a charging capacitor. Consider an Amperian loop encircling the wire that feeds one plate.

If you cap that loop with a flat surface cut by the wire, a real conduction current $I_c$ passes through it.
If you instead cap the SAME loop with a bulging surface that passes between the two capacitor plates, no charge crosses the gap — the space between the plates is an insulator. So $I_c = 0$ through this surface.

Ampere's law now gives two different answers for $\oint \vec{B}\cdot d\vec{l}$ for the same loop, depending on which surface you choose. That is a contradiction — the law cannot be right as it stands.

Maxwell's Fix: A Current Made of Changing Field

Between the plates there is no moving charge, but there is a growing electric field, because charge is piling up on the plates. Maxwell proposed that a changing electric flux acts like a current for the purpose of producing a magnetic field. He called it the displacement current, $I_d$ .

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

where $\Phi_E = \int \vec{E}\cdot d\vec{A}$ is the electric flux through the surface, and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space.

Check with the capacitor. For a parallel-plate capacitor of area $A$ and plate charge $q$ , the field between the plates is $E = \dfrac{q}{\varepsilon_0 A}$ , so the flux is $\Phi_E = EA = \dfrac{q}{\varepsilon_0}$ . Then

I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \cdot \frac{1}{\varepsilon_0}\frac{dq}{dt} = \frac{dq}{dt} = I_c

So the displacement current in the gap is exactly equal to the conduction current in the wire. The two surfaces now give the same answer — the contradiction is gone.

The Complete (Ampere–Maxwell) Law

Maxwell rewrote Ampere's law so that the total current is conduction plus displacement current:

\oint \vec{B} \cdot d\vec{l} = \mu_0\left(I_c + I_d\right) = \mu_0 I_c + \mu_0\varepsilon_0 \frac{d\Phi_E}{dt}

Important

The deep meaning: a changing electric field produces a magnetic field, just as (by Faraday's law) a changing magnetic field produces an electric field. This symmetry is what makes self-sustaining electromagnetic waves possible — the changing E-field of the wave generates the B-field and vice versa.

Key Points to Remember …

Why this formula?

Displacement Current: Why the Formula Holds

The displacement current is one of the most elegant corrections in physics — it fixed a logical flaw in Maxwell's equations and predicted electromagnetic waves. Let's understand why its formula emerges.

1. The Problem That Demanded a Fix

Consider a capacitor being charged in a circuit. Ampère's law (in its original form) states:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

where $I_{\text{enc}}$ is the current passing through any surface bounded by the loop.

Now take two different surfaces bounded by the same loop:

Surface S₁: Cuts the wire — current $I$ passes through.
Surface S₂: Passes between the capacitor plates — no current passes through.

Surface	Current through it
S₁ (cuts wire)	$I$
S₂ (between plates)	$0$

This is a contradiction: the same loop gives two different values for $\oint \vec{B} \cdot d\vec{l}$ . Ampère's law is inconsistent for time-varying fields.

2. The Insight: Changing Electric Field

Between the capacitor plates, there is no conduction current, but there is a changing electric field as charge builds up.

The electric field between plates: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$
As $Q$ changes, $E$ changes: $\frac{dE}{dt} = \frac{1}{\varepsilon_0 A} \frac{dQ}{dt}$

Maxwell realized: a changing electric field should produce a magnetic field, just like a current does.

3. Deriving the Displacement Current Formula

Step 1: Relate charge to electric flux

The electric flux through the capacitor plates is:

\Phi_E = \int \vec{E} \cdot d\vec{A} = E \cdot A = \frac{Q}{\varepsilon_0}

Step 2: Differentiate with respect to time

\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dQ}{dt} = \frac{I}{\varepsilon_0}

Step 3: Define displacement current

Maxwell defined the displacement current $I_d$ as:

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

From Step 2, this equals $I$ — the same conduction current in the wire. The displacement current "bridges" the gap.

4. The Corrected Ampère-Maxwell Law

The full law becomes:

\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_{\text{enc}} + I_d)

Or equivalently:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}

Why this works:

For surface S₁: $I_{\text{enc}} = I$ , $\frac{d\Phi_E}{dt} = 0$ → result = $\mu_0 I$
For surface S₂: $I_{\text{enc}} = 0$ , $\frac{d\Phi_E}{dt} = \frac{I}{\varepsilon_0}$ → result = $\mu_0 \varepsilon_0 \cdot \frac{I}{\varepsilon_0} = \mu_0 I$

Both surfaces give the same answer. The contradiction is resolved.

5. The Key Formula(e) — Summarized

Quantity	Formula	Meaning
Displacement current	$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$	Equivalent "current" from changing E-field

Start your 14-day free trial to unlock the full solution →

The displacement current through the capacitor is exactly equal to the conduction current in the wires at all times — it is $I_d = -2\pi\nu q_0 \sin(2\pi\nu t)$ , which is the same as $dq/dt$ for the given charge variation.

Why Displacement Current?

When a capacitor charges or discharges, no actual charge flows through the dielectric between the plates. Yet the magnetic field around the wires and between the plates behaves as if a current were flowing. Maxwell resolved this paradox by introducing the displacement current — a term that accounts for the changing electric field between the plates.

The key insight: wherever the electric flux $\Phi_E$ changes with time, there is an equivalent "current" given by:

I_d = \epsilon_0 \frac{d\Phi_E}{dt}

For a parallel plate capacitor, the electric field between the plates is uniform (neglecting edge effects) and directly related to the charge on the plates. So the displacement current becomes simply the rate of change of charge — the same as the conduction current feeding the capacitor.

Let's work through it step by step.

The electric field between the plates

For a parallel plate capacitor with plate area $A$ and charge $q$ , the electric field between the plates (neglecting edge effects) is:

E = \frac{\sigma}{\epsilon_0} = \frac{q}{\epsilon_0 A}

This field is uniform and perpendicular to the plates.

Electric flux through a surface between the plates

Consider any surface parallel to the plates, of area $A$ . The electric flux through it is:

\Phi_E = E \cdot A = \frac{q}{\epsilon_0 A} \cdot A = \frac{q}{\epsilon_0}

Notice that the area cancels — the flux depends only on the charge, not on the plate area. This is a direct consequence of Gauss's law.

Displacement current definition

The displacement current through any surface is:

I_d = \epsilon_0 \frac{d\Phi_E}{dt}

Substituting $\Phi_E = q/\epsilon_0$ :

I_d = \epsilon_0 \frac{d}{dt}\left(\frac{q}{\epsilon_0}\right) = \frac{dq}{dt}

This is the central result: the displacement current between the plates equals the conduction current in the wires.

Tip

This equality is not a coincidence — it's exactly what Maxwell needed to make the total current continuous around the circuit. The displacement current "completes" the circuit through the capacitor gap.

Apply the given charge variation

We are given $q = q_0 \cos(2\pi\nu t)$ . Differentiate:

\frac{dq}{dt} = -q_0 \cdot 2\pi\nu \cdot \sin(2\pi\nu t)

$= -2\pi\nu q_0 \sin(2\pi\nu t)$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$Q. Which of the following is incorrect for displacement current A. Displacement current exists only when electric field changes with time B. Displacement current obeys ohm's law C. Displacement current produces magnetic field just like conduction current D. Unlike conduction current, displacement current does not involve the flow of electrons through a conductor (A) B (B) D (C) A (D) C$
›Reveal solutionSolution
Displacement current does not obey Ohm's law, so statement B is the false one; the option that names statement B is option (A).

Concept

Displacement current, introduced by Maxwell to complete Ampère's law, has density $J_d=\varepsilon_0\dfrac{\partial E}{\partial t}$ . It is not a flow of charge: it arises purely from a changing electric field and, like a real current, produces a magnetic field. Because it carries no charge carriers and no resistance, it has no $J=\sigma E$ (Ohm's-law) relation.

Solution

Test each statement:

A — "exists only when $E$ changes with time": true, since $J_d\propto \partial E/\partial t$ ; a static field gives zero displacement current.

B — "obeys Ohm's law": false. Ohm's law $J=\sigma E$ applies to conduction currents; displacement current depends on $\partial E/\partial t$ , not on $E$ through a conductivity.

C — "produces a magnetic field just like conduction current": true — this is precisely Maxwell's correction, $\nabla\times B=\mu_0\big(J_c+\varepsilon_0\partial E/\partial t\big)$ . …
Start your 14-day free trial to see the full solution →
KCET 2024Set D-21 markMCQ
$5\,\mu\text{F}$
›Reveal solutionSolution

Differentiate the stored-energy formula $U=\tfrac12 CV^2$ with respect to time and substitute the instantaneous values.

Step 1 — The concept: energy stored in a capacitor

At any instant when the potential difference across a capacitor is $V$ ,
$U = \frac{1}{2}CV^2$
Note this is the instantaneous energy — it depends on the capacitor's present voltage $V$ , not on the battery's emf. The emf of $10\ \text{V}$ only tells us the capacitor is still charging (it has reached $4\ \text{V}$ of an eventual $10\ \text{V}$ ).

Step 2 — Differentiate with respect to time

$C$ is a constant of the capacitor, and $V$ is the only time-dependent quantity, so by the chain rule:
$\frac{dU}{dt} = \frac{1}{2}C \cdot 2V\,\frac{dV}{dt} = C\,V\,\frac{dV}{dt}$
Step 3 — Substitute the instantaneous values
$C = 5\ \mu\text{F} = 5\times10^{-6}\ \text{F},\qquad V = 4\ \text{V},\qquad \frac{dV}{dt} = 0.6\ \text{V\,s}^{-1}$
$\frac{dU}{dt} = \left(5\times10^{-6}\right)(4)(0.6)$ …
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-M1 markMCQ
$400 \mathrm{~pF}$
›Reveal solutionSolution
In a capacitor connected to an AC source, the displacement current between the plates equals the conduction current in the wires at every instant. Given the rms current is $4\ \mu\text{A}$ , the displacement current is also $4\ \mu\text{A}$ . The correct option is (C).

The key idea is that displacement current is not a separate physical current; it is the term Maxwell added to make the total current continuous in circuits containing capacitors. In a parallel plate capacitor, the displacement current between the plates exactly equals the conduction current in the connecting wires at every moment. This is because the changing electric field between the plates produces a displacement current density $\mathbf{J}_d = \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$ , and integrating over the plate area gives the same current as flows into the capacitor.

Thus, if the rms conduction current in the wire is $4\ \mu\text{A}$ , the rms displacement current between the plates is also $4\ \mu\text{A}$ . No further calculation is needed — the problem tests this conceptual equality.

Let’s verify with the given numbers to be thorough:

Given data

Capacitance: $C = 400\ \text{pF} = 400 \times 10^{-12}\ \text{F}$

AC source voltage (rms): $V_{\text{rms}} = 100\ \text{V}$

Angular frequency: $\omega = 100\ \text{rad/s}$

RMS current: $I_{\text{rms}} = 4\ \mu\text{A} = 4 \times 10^{-6}\ \text{A}$

Check consistency

For a capacitor, the impedance is $X_C = \frac{1}{\omega C}$ .

$X_C = \frac{1}{100 \times 400 \times 10^{-12}} = \frac{1}{4 \times 10^{-8}} = 2.5 \times 10^{7}\ \Omega$
The rms current from Ohm’s law for AC would be:

$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{100}{2.5 \times 10^{7}} = 4 \times 10^{-6}\ \text{A} = 4\ \mu\text{A}$ …
Start your 14-day free trial to see the full solution →
KCET 2021Set B-21 markMCQ
$Q. The source of electromagnetic waves can be a charge. (A) Moving with a constant velocity (B) Moving in a circular orbit (C) At rest (D) Moving parallel to the magnetic field$
›Reveal solutionSolution
A charge emits electromagnetic waves only when it accelerates. Among the given options, only circular motion involves acceleration (centripetal acceleration), so the correct choice is (B).

The key idea is simple: electromagnetic waves are produced by accelerating charges. A charge at rest or moving with constant velocity (in a straight line) produces only static or steady fields — no radiation. But when a charge accelerates, its electric and magnetic fields change in time, and those changes propagate outward as electromagnetic waves.

Let’s examine each option carefully.

Option (A): Moving with a constant velocity

A charge moving at constant velocity has zero acceleration. Its electric field is steady (though it moves with the charge), and its magnetic field is also steady. No time-varying fields are produced, so no electromagnetic waves are emitted.

Watch out

A common mistake is to think that any moving charge radiates. Only accelerating charges radiate — constant velocity is not enough.

Option (B): Moving in a circular orbit

Circular motion, even at constant speed, involves continuous acceleration toward the centre (centripetal acceleration). This acceleration causes the charge’s velocity direction to change constantly, producing time-varying electric and magnetic fields. These fields propagate away as electromagnetic waves. This is exactly how synchrotron radiation is produced — electrons in a circular accelerator emit intense EM waves.

Option (C): At rest …
Start your 14-day free trial to see the full solution →
KCET 2019Set A-11 markMCQ
$\frac{R^2 + \omega^2 L^2}{V}$
›Reveal solutionSolution
In an L-R series AC circuit, the average power dissipated is $P = V_{\text{rms}} I_{\text{rms}} \cos\phi = \frac{V^2 R}{R^2 + \omega^2 L^2}$ , which matches option (B).

The key idea is that in an AC circuit, power is not simply $V^2/R$ because the inductor stores and returns energy without dissipation. Only the resistor consumes real power, and the phase difference between voltage and current reduces the net power.

For a series L-R circuit driven by an AC source of rms voltage $V$ and angular frequency $\omega$ , the impedance is $Z = \sqrt{R^2 + (\omega L)^2}$ . The rms current is $I = V/Z$ . The power factor is $\cos\phi = R/Z$ , where $\phi$ is the phase angle by which current lags voltage. Average power dissipated is $P = V I \cos\phi$ .

Write the impedance:

$Z = \sqrt{R^2 + (\omega L)^2}$

The rms current:

$I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (\omega L)^2}}$

The power factor:

$\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (\omega L)^2}}$

Average power: …
Start your 14-day free trial to see the full solution →
KCET 2018Set A-11 markMCQ
$100~\mathrm{W}$
›Reveal solutionSolution

A filament bulb is a pure resistor, so $P = V_{rms}I_{rms}$ with power factor $1$ ; simply divide.

Step 1 — Why the formula is just $P = VI$ .

In a general AC circuit the average power is
$P = V_{rms}\, I_{rms}\, \cos\phi$
where $\cos\phi$ is the power factor. A bulb's filament is a pure resistance — no inductance or capacitance — so voltage and current are in phase, $\phi = 0$ and $\cos\phi = 1$ . The $50~\mathrm{Hz}$ therefore plays no role at all; it is given only as a distractor.

Step 2 — Substitute the ratings.

Ratings on a bulb are always rms (average) values.

$I_{rms} = \frac{P}{V_{rms}} = \frac{100\ \text{W}}{220\ \text{V}}$ …
Start your 14-day free trial to see the full solution →

←Previous questionQ9 — Why does microwave oven heats up a food item containing water molecule…→Next questionQ11 — A variable frequency a.c. source is connected to a capacitor. How will…

Q.The charge on a parallel plate capacitor varies as $q = q_0\cos 2\pi\nu t$ . The plates are very large and close together (area $= A$ , separation $= d$ ). Neglecting the edge effects, find the displacement current through the capacitor.

Concept understanding — Displacement Current

Displacement Current

The Problem Maxwell Spotted

Maxwell's Fix: A Current Made of Changing Field

The Complete (Ampere–Maxwell) Law

Key Points to Remember …

Why this formula?

Displacement Current: Why the Formula Holds

1. The Problem That Demanded a Fix

2. The Insight: Changing Electric Field

3. Deriving the Displacement Current Formula

Step 1: Relate charge to electric flux

Step 2: Differentiate with respect to time

Step 3: Define displacement current

4. The Corrected Ampère-Maxwell Law

5. The Key Formula(e) — Summarized

Why Displacement Current?

Method: Computing Displacement Current from a Given Time-Varying Charge

Steps

Concept

Solution

Step 1 — The concept: energy stored in a capacitor

Step 2 — Differentiate with respect to time

Step 3 — Substitute the instantaneous values

Q.The charge on a parallel plate capacitor varies as q=q0cos⁡2πνtq = q_0\cos 2\pi\nu tq=q0​cos2πνt. The plates are very large and close together (area =A= A=A, separation =d= d=d). Neglecting the edge effects, find the displacement current through the capacitor.

Why Displacement Current?

🎓Unlock everything free for 14 days

Q.The charge on a parallel plate capacitor varies as $q = q_0\cos 2\pi\nu t$ . The plates are very large and close together (area $= A$ , separation $= d$ ). Neglecting the edge effects, find the displacement current through the capacitor.

Unlock everything free for 14 days